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Load 1 is the ignitor.
Yes, everything is basically “connected”, like in any configuration of parallel and/or series circuits. But can the electrons get from the ignitor to L1 and N without having to go through another load? Yes or no.
Remember – if they have a direct path along wire, they will take it.
For now just analyze this with the detector switch being closed, essentially making that branch a wire. We’ll talk about if it were open later…
Hi Matt,
The portion of the webinar referred to in this thread was not recorded (free-form Q&A parts usually aren’t).
All of the concepts you need to know to answer this question were covered in the module, mostly in unit 5. But this diagram is drawn a little oddly, which is one reason why we use it. You’ll encounter all kinds of variations when you look at schematics, so we want our students to have to stretch a little to help them gain experience and confidence!
You are actually making the circuits more complicated than they are. After all, if you were going to have to divide up the voltage drop across multiple loads, you would have to know something about their relative resistances, right? And we don’t give that in the problem statement. So step back and take another look at the diagram.
Become each load, one at a time. Any of the loads that can reach out and touch L1 and N without having to go through another load is by itself in a circuit, in parallel with the others. Don’t get fooled by how the lines are drawn – even if there is a bend or a junction, as long as the electrons can travel along a wire rather than through another load, they will.
Let me know for each load what you see when you do that Zen trick. (see the video in this unit for a refresher https://my.mastersamuraitech.com/module-3/basic-electricity-series-and-parallel-circuits/)
Hi Steve,
Read through this Forum thread on Req and see if that helps you figure it out.
Let me know if you still have questions after looking though this: Calculating Equivalent Resistance
Hi Steve,
This is such a good question that I decided to add some material to Basic Electricity, unit 3, to help clarify using Ohm’s Law and specifically why using P = E2/R doesn’t work where E is the supplied voltage and R is the resistance of the loose connection. (you can look at that here: https://my.mastersamuraitech.com/module-3/basic-electricity-ohms-law/)
It has to do with the fact that in a series circuit, supply voltage will drop across the loads in proportion to the resistance of each load. The total of these voltage drops will equal the voltage supply. So, if you have two loads in this oven circuit, neither one has a voltage drop of 240 – the voltage drops will add up to 240.
We teach a lot more about voltage drop in the subsequent units, so don’t think you have to completely understand it at this moment. But it’s an important concept, so pay attention as we continue to talk about it!
I hope that helps. Let me know if you have any follow-up questions.
Hi Mario,
If you look at the diagram again, the reading across L1 and L2 is zero. Each end of the element with respect to N is 120vac.
First, let me review what voltage drop is. When current flows through a load (resistance), a voltage drop across that load will be created (E = I x R).
The video states that the element has continuity, so the fact that there’s no voltage drop means there is no current flowing through the circuit. Something is open.
The zero reading across the load also means that these are electrically equivalent points (EEPs). It’s like reading two points along a wire.
So what about those L1 to N and L2 to N readings? Those are voltage potential difference readings (not voltage drop), and now that we know that the element is just acting like a wire, and the ends of the element are EEPs, what can we determine? Are we really reading two different voltage differences?
Lemme know what you think …
Hi John,
Have you seen the newer videos we’ve added to the Basic Electricity module since you first went through it? There are some good ones, particularly in units 4, 5, and 8 that might help you understand voltage drop better.
But the short answer is that in a series circuit, the total voltage dropped across all of the loads will equal the source voltage. Or, put another way, if you take the voltage drop across each load in a series circuit and add them together, the sum will equal the source voltage.
The amount of voltage drop across each load depends on the resistance of that load. So, if one load is very high R and the other very low R, then most of the voltage will be dropped across the high R load.
As to any specific question, such as the ice maker, we’d have to see the schematic to be able to help you on that. I believe you’ve also posted about this over at Appliantology – so one place or another, you can post the schematic if you want more help.
Hi Brian,
You are correct! (I removed the numerical answer you gave, just so we don’t give it away to other students who happen to see this.)
Current is the same at every point in a series circuit, so the current has to be calculated using the total resistance in the circuit. Once you know the current, you can use the Ohm’s Law equation for Power along with the resistance of the loose connection to calculate the heat generated.
🙂
Hi Joe,
If you are asking this because you have another version of the book, such as on Kindle, here’s a pdf version of the book you can use to look up by page number:
To answer your question – we don’t have a list of the corresponding topic names.
Wire colors don’t always follow convention, so you have to trace them to know what’s what.
Actually, I have a follow-up!
What we need a load to do is “work”, right? This means we primarily need a certain amount of power (voltage AND current) for these components to operate, P=IxE, or P=I^2xR are two common ways to calculate that. Loads run on power (watts), not voltage alone.
When a component has a voltage rating, it means that in order to get the needed current flow to create the Power it needs to operate, it must drop that rated amount of voltage to achieve that.
So, just to reiterate based on your question, a component rated at 120v would need to drop right around 120v to operate according to spec, thus it could not be in a circuit (AC or DC) that contains another 120v load, unless it is a 240v circuit.
Hi Josh,
Good questions!
I’m going to answer your last question first: Ohm’s Law works the same for both AC and DC voltages. The fact that AC is switching polarity doesn’t affect these basic calculations that we’re talking about.
When you have loads in series, the combined voltage drop will equal the voltage supply. So, you are correct when you said that two loads of the same resistance in series will drop 60v each to total 120v. (In a 120v circuit)
You refer to a “120v door lock” or “120v drain pump”. The “120v” is the rating for that component – the voltage it should be getting to function properly.
If two components both need 120v, would they be placed in series in a 120v circuit? No. They would have to be a different configuration – parallel circuits, or use of switches and shunts – to ensure that they are getting their rated voltage.
Does that make sense? Any more follow up questions?
Hi Colin,
It is possible, but not something we advise doing since it’s not always saving that much money and it can be tricky to determine the exact state of all the various components. The best practice is to replace the board.
Hi Brian,
Your original answer (down) is the correct one! I wonder if you accidentally looked at the wrong question/answer the first time you took it? I double-checked the quiz and it still is set up correctly.
But, your original thinking is correct, that’s the important part!
And by the way, we have lots of students who are completely new to the field! You’ll do fine – it may take a little more work or repeating of some lessons than for those who already have some experience, but if you put in the work you’ll learn what you need to!
And feel free to ask questions when you need to! Just please start topics in the Forum for that particular course and unit. Thanks!
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